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Facoltà di Scienze Matematiche, Fisiche e Naturali

Corso di Laurea Magistrale in Matematica

GALOIS MODULE STRUCTURE OF THE

SQUARE ROOT OF THE

INVERSE DIFFERENT

Tesi di Laurea Magistrale

27 Ottobre 2017

Relatore

Prof.ssa Ilaria Del Corso

Controrelatore

Candidato

Prof. Roberto Dvornicich

Matteo Verzobio

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Contents

1 Square root of the inverse dierent 1

1.1 Orders and maximal orders . . . 1

1.2 The generalized idele group . . . 4

1.3 The classgroup . . . 5

1.4 Hom-description . . . 8

1.5 Tate's cohomology . . . 12

1.6 Square root of the inverse dierent . . . 15

2 The odd degree case 21 2.1 Localization to a maximal order . . . 21

2.2 Proof of Theorem 2.1.10 in tame case . . . 25

2.3 Proof of Theorem 2.1.10 in wild case . . . 29

2.4 Conclusion of odd degree case . . . 33

3 The even degree case 35 3.1 Reduction to local case . . . 35

3.2 The ideal M . . . 47

3.3 The content of s . . . 50

3.4 Stickelberger and Hasse-Davenport Theorems . . . 54

3.5 The class of S . . . 58

3.6 Sympletic and quaternionic characters . . . 59

3.7 The even degree case . . . 64

3.8 Counterexamples . . . 68

Bibliography 77

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Introduction

Let G be a nite group and dene the classgroup of the locally free modules over Z[G] as the group of the classes of locally free Z[G]-modules with the equivalence relation

M ∼ N ⇔ ∃n, m ∈ N s.t. MMZ[G]m∼= N M

Z[G]n

and the operation given by the direct sum of modules. In this thesis we will consider the class group in the particular case when G is the Galois group of a Galois extension of number elds N/E. The purpose of this work is to study the Z[G]-modules which appears in this context, as elements of Cl(Z[G]). In the case when N/E is a tame extension it is known that ON (the ring of integers of N) is

locally free and it class has been deeply investigated. By Taylor's Theorem [27], this class can be expressed in terms of the Artin root number. In this thesis we consider, when it exists, the square root A(N/E) of the inverse of the dierent C(N/E) = {x ∈ N |TrN/E(xON) ⊆ OE}. In the case of odd degree extensions,

A(N/E)always exists and is locally free over Z[G] if and only if N/E is weakly ramied. This context has been studied by Erez [7], who proved

Theorem 1. Let N/E be a tame, odd degree Galois extension of number elds with Galois group G. Then A(N/E) is free over Z[G].

This result follows from an other important theorem of Erez: for every Z-maximal order M in Q[G] which contains Z[G], the M-module A(N/E)⊗Z[G]M

is free. The most important instrument involved in this work is a theorem of Fröhlich, known as the Hom-Description, which gives an isomorphism between the classgroup and a quotient of the group of the homomorphisms from additive group of virtual characters of G to the idele group of a number eld. Since Taylor's Theorem implies the triviality of (ON)when N/E is tame and has odd

degree, we get (A(N/E)) = (ON)and both classes are in fact trivial (we denote

with (M) the class of a locally free module M in Cl(Z[G])).

It is natural to consider the problem to determine the class of A(N/E) also when the extension is tame and the degree is even. Assuming the existence of A(N/E) (which is not guaranteed in general) we can prove that, if N/E is tame, then A(N/E)is locally free over Z[G]. This case has been studied by Caputo and Vinatier in [1], where some generalizations of the result of Erez are presented. The fundamental result which we present in this thesis, is the following Theorem 2. Let N/E be a tame locally abelian Galois extension of number elds. Assume further that A(N/E) exists. Then

(A(N/E)) = (ON) ∈Cl(Z[G]).

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We extend the denition of Cl(Z[G]) to the G-cohomologically trivial modules and we introduce the torsion module S := A(N/E)/ON, which turns out to be

G-cohomologically trivial, hence it dene a class in Cl(Z[G]). A fundamental step of the proof is to show that the class of S is trivial; the principal tools for the proof are the Stickelberger theorem, Hasse-Davenport theorem and the study of quaternionic characters.

As far as the triviality of the class of A(N/E) is concerned, the following theorem holds:

Theorem 3. [1] Let N/E be a tame locally abelian Galois extension whose inverse dierent is a square and suppose that no archimedean places ramies in N. Then

(A(N/E)) = 1 ∈Cl(Z[G]).

This results follows from the study of the Artin root number of sympletic char-acters, combined with Taylor's Theorem. However, the class of A(N/E) is not always trivial, even for tame locally abelian extensions. In fact, Caputo and Vinatier, exhibit an example of tame locally abelian Galois extension of number elds whose inverse dierent is a square such that

(A(N/E)) 6= 1 ∈Cl(Z[G]).

We explicitly describe a tame locally abelian SL2(F3)-Galois extension N/Q

such that (A(N/Q) 6= 1 in Cl(Z[SL2(F3)]) and we show that this example is

minimal in the sense that (A(N/Q)) = 1 if N/Q is a tame locally abelian G-Galois extension with cardG < 24.

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Chapter 1

Square root of the inverse

dierent

In this chapter we want to introduce the objects and the instruments involved in this work. We will dene the orders and the classgroup of the locally free modules over an order, i.e, a free group generated by locally free modules with an equivalence relation. Afterward, we will introduce the instruments to study the classgroup: the Tate's cohomology, the generalized idele group and the Hom-description, which give us a very useful isomorphism. Finally, we will introduce the main subject of this work, the square root of the inverse dierent of a Galois extension of number elds, endowed with a structure of Z[G]-module with G the Galois group of the extension. We will try to understand when it is locally free over Z[G] and, in this case, we will study it in the classgroup of Z[G].

1.1 Orders and maximal orders

If K is a number eld, one can associate to K its ring of integers OK. To

generalize the concept of ring of integers to a K-algebra, we dene the orders. Let R be a noetherian integral domain with quotient eld K.

Denition 1.1.1. For any nite dimensional K-space V a full R lattice in V is a nitely generated R-module M in V such that KM = V .

Denition 1.1.2. Let A be a K-algebra. Λ is an R-order in A if • Λ is a subring of A.

• Λ has the same unity of A. • Λ is a full R lattice in A.

Example 1.1.3. If G is a nite group, then K[G] is a nite dimensional K-space. R[G] is an R-order because it is a subring of K[G], it is a nitely generated R-module and KR[G] = K[G].

Theorem 1.1.4. Every element of an R-order Λ in A is integral over R. 1

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Proof. For each a ∈ Λ, it follows from the inclusion R[a] ⊆ Λ that R[a] is a nitely generated R-module (since R is noetherian), then a is integral over R.

Denition 1.1.5. A maximal order is an order that is not contained properly in any other order.

Corollary 1.1.6. If the integral closure of R in A is an order, then the integral closure is the unique maximal order.

Proposition 1.1.7. Every R-order in A is contained in a maximal R-order in A. There exists at least one maximal R-order in A.

Proof. See [5, p.127].

Proposition 1.1.8. If G = Z/nZ, then Q[G] has a unique maximal Z-order. Proof. Let us denote with φn(x)the n-th cyclotomic polynomial. Then,

Q[G]∼= Q[x]xn− 1 ∼= Q[x] Y d|n φd(x) ∼= Y d|n Q(ζd).

It is easy to check that every maximal order of Q[G] goes in a maximal order of Qd|nQ(ζd) with the isomorphism above and then we just need to check

that Qd|nQ(ζd) has only one Z maximal order. The integral closure of Z in

the product is the product of the closures. Then, the integral closure of Z is Q

d|nZ[ζd]. This is an order and so we have that Qd|nZ[ζd] is the maximal

order. We can observe that in general is not true that Z[G] is the maximal order since in general

Z[x] Y d|n φd(x) 6∼= Y d|n Z[x]φd(x).

Proposition 1.1.9. If G is nite and abelian, then Q[G] has a unique maximal Z-order.

Proof. Suppose G ∼= Z/n1Z × Z/n2Z. Then, Q[G]∼=Q[x, y]/(xn1− 1)(yn2− 1) ∼ =  Q[x]/(xn1− 1)  ⊗QQ[y]/(yn2− 1)  ∼ = Y d1|n1 d2|n2  Q[ζd1] ⊗QQ[ζd2]  .

This is a product of elds because every term in the product is a eld or a product of elds. Then, we conclude as in the previous proposition. In general for the classication of nite abelian group we have that

G ∼= Z/n1Z × Z/n2Z · · · × Z/nkZ and then

Q[G]∼= Y

di|ni

Q[ζd1] ⊗ · · · ⊗ Q[ζdk]

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Lemma 1.1.10. Let R be a noetherian integral domain with quotient eld K and let G be a nite group of order n. If Λ is an order that contains R[G], then

R[G] ⊆ Λ ⊆ n−1R[G].

Proof. First of all, we recall the denition of the characteristic polynomial. If α ∈ K[G]and {v1, . . . , vn}is a basis of K[G] over K, then we have

αvi=

X ai,jvj

and

charpol(α)(x) = det(δi,jx − ai,j).

Observe that the coecient of xn−1 is (− P

iai,i). For [5, Proposition 1.9], α

is integral over R if and only if charpol(α) ∈ R[x]. We dene the trace of an element in K[G] as the K-linear map such that

T (g) = (

0 if g 6= e

nif g = e. (1.1) If we take {g}g∈G as a basis of K[G] over K, it is easy to check that −T (α) is

the coecient of xn−1of the characteristic polynomial of α. Let

λ =Xagg ∈ Λ

with ag ∈ K. Then, for every γ ∈ G, we have that γ−1λ ∈ Λ (because Λ is a

ring and R[G] ⊆ Λ) and then

T (γ−1λ) ∈ R,

since every element of Λ is integral over R. Thus aγ ∈ R/nsince T (γ−1λ) = naγ

and the searched inclusion follows.

Theorem 1.1.11. Let R be a noetherian integral domain with quotient eld K and let G be a nite group of order n. Then, R[G] is a maximal R-order of K[G]if and only if n−1∈ R.

Proof. If n−1∈ Rand Λ is an order which contains R[G], then

R[G] ⊆ Λ ⊆ n−1R[G] = R[G]

and we conclude that R[G] is a maximal order. Vice versa, let us consider the element z = 1 n X g∈G g;

we note that it belongs to the center of the algebra K[G] and it is idempotent. Since gz = z for every g ∈ G, it follows that R[G] + Rz is an order and then z ∈ R[G], which implies n−1∈ R.

Corollary 1.1.12. If G is a nite group of order n, then the Zp-order Zp[G]of

Qp[G]is maximal if and only if (p, n) = 1.

Proposition 1.1.13. If K is a number eld, then OK is the maximal order of

the K-algebra K. Proof. Trivial for 1.1.6.

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1.2 The generalized idele group

In this section we want to generalize the concept of idele group. As we noted above, orders are, in some sense, a generalization of the rings of integers; pur-suing in this analogy, we can dene the idele group associated to an order, generalizing the classical idele group, which is associated to the ring of integers. First of all, we recall the classical denition of adele ring and idele group. Proposition 1.2.1. Let K be a number eld and let P be a place of K. We denote with KP the completion of K with respect to P .

Denition 1.2.2. The adele ring of a number eld K is the ring VK= {α = (αP) ∈

Y

KP | αP ∈ AP almost everywhere},

where the product runs over all the places of K and AP = OKP if P is nite,

AP = KP otherwise.

The ring VK has a topological structure given by the basis of open sets Q ΓP

with ΓP open sets of AP for every P and ΓP = AP a.e.

Denition 1.2.3. Consider the embedding ι : VK∗ ,→ VK× VK

dened by

x → (x, x−1). The idele group is V∗

K endow with the topology of subspace induced by the

previous inclusion. That means, on ι(V∗

K) we consider the subset topology of

the product topology. The group operation is the point-wise multiplication. Consider now the case of K-algebras. Let A be a K-algebra and let Λ be an OK-order. Let us dene

AP = A ⊗KKP and ΛP = Λ ⊗OKOKP.

We dene the generalized idele group as J (A, Λ) = {α = (αP) ∈

Y

AP | αP ∈ Λ∗P a.e. if P is nite}

where the product runs over all places. Observe that if Λ and Λ0are two dierent

orders then there exists m ∈ Z such that

mΛ0 ⊆ Λ ⊆ (1/m)Λ0.

It follows that, if P is a prime of OK above a rational prime that does not

divide m, then ΛP = Λ0P. Since there is only a nite number of prime dividing

m, the idele group does not actually depend on the particular choice of the order. Finally let us dene

U (Λ) =YΛ∗P ⊆ J (A, Λ).

Example 1.2.4. Let G be a nite group. Then, Z[G] is a Z-order of the Q-algebra Q[G]. For every nite prime p we have (Q[G])p = Qp[G]and (Z[G])p= Zp[G].

So

J (Q[G], Z[G]) = {α = (αP) ∈

Y

P places

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1.3 The classgroup

In this section we want to give an introduction to the classgroup. For a more detailed presentation see [9] and [6, Chapter 6].

Let K be a number eld, G a nite group, Λ an OK-order in K[G] and M a

nitely generated module over Λ. For every place P of K we set: MP = M ⊗OKOKP and ΛP = Λ ⊗OKOKP.

Clearly, MP is a ΛP-module.

Denition 1.3.1. We say that M is locally free if, for every prime P of OK,

the module MP is free over ΛP.

Proposition 1.3.2. Let M be a locally free module over Λ. The rank of M is dened as the rank of the free K[G]-module M ⊗OKK. This rank is nite and

it is also the rank of MP over ΛP for all primes P .

Proof. See [8, 1.4].

Let Λ be an OK-order of K[G]. We dene K0(Λ)as the abelian group generated

by locally free modules over Λ with the relations [M ] + [N ] = [M ⊕ N ].

It appears clear that every element in K0(Λ)can be written in the form [M −N]

with M and N locally free. Consider the homomorphism f : Z → K0(Λ)

that sends every n ∈ N in Λn and −n in −Λn.

Denition 1.3.3. We dene the classgroup as Cl(Λ) := Coker(f).

Let us observe that a locally free module has trivial class in the classgroup if [M ] = [Λr− Λs], that is equivalent to M ⊕ Λs= Λr.

We can also give an equivalent denition of the classgroup. Consider the homo-morphism

Rk : K0(Λ) → Z

dened by

[M − N ] →RkM − RkN, where RkM is the rank map of Proposition 1.3.2. Then,

Cl(Λ) ∼= ker(Rk)

where the isomorphism between Cl(Λ) and ker(Rk) is given by M → M − Λm,

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Denition 1.3.4. Let A be a nitely generated K-algebra. We say that A is simple if it contains no non-trivial two-sided ideals and the multiplication operation is not zero (that is, there is some a and some b such that ab 6= 0). We say that A is semisimple if A = Q Ai with Ai simple algebras.

Theorem 1.3.5 (Maschke). If G is a nite group and K is a eld of charac-teristic 0, then K[G] is a semisimple algebra.

Proof. See [15, p.666]

Recall that, given B a simple algebra with its center C, we can dene a C-linear map φb(x) = bxfrom B to itself.

Denition 1.3.6. The reduced norm map from B to C is a map that asso-ciates to an element b the constant term of the characteristic polynomial of φb,

multiplied for (−1)m, with m the dimension of B over C.

Let A be a semisimple algebra with center L. If Q Ai is the decomposition of

A into simple algebras, then L = Q Li with Li the center of Ai. The reduced

norm

n : A 7→ L

is the map whose components are the reduced norms of its simple factors. This map can be extended to

n : J (A, Λ) 7→ J (L, Λ). We dene J(A, Λ)0 as the ker of n.

Proposition 1.3.7. Let V be a left K[G]-module, and M an OK full lattice in

V. If α ∈ J(K[G], Λ), then there is a unique full OK lattice I in V such that

IP = αPMP

for all P . We shall denote I with αM. Proof. See [9, Theorem 1].

Theorem 1.3.8. Every locally free module over Λ of positive rank m is iso-morphic to a direct sum

α1Λ ⊕ · · · ⊕ αmΛ

with αi∈ J (K[G], Λ). Moreover, if

α1Λ ⊕ · · · ⊕ αmΛ ∼= β1Λ ⊕ · · · ⊕ βnΛ

then necessarily n = m and if m = 1, then

(K[G])∗α1U (Λ) = (K[G])∗β1U (Λ)

and if m > 1 we have

(α1α2...αm)J (K[G], Λ)0(K[G])∗U (Λ) = (β1β2...βm)J (K[G], Λ)0(K[G])∗U (Λ).

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Corollary 1.3.9. Every locally free right Λ-module of positive rank is isomor-phic to a direct sum of free modules and a right ideal of Λ.

Proof. See [9, p.115].

Theorem 1.3.10. Using the previous notation Cl(Λ) ∼= J (K[G], Λ)

J (K[G], Λ)0K[G]U (Λ).

Proof. See [6, Theorem 49.22.]

Denition 1.3.11. We say that the order Λ has the cancellation property, if, for all pairs, M and N, of locally free modules over the ring Λ, it holds:

M ⊕ Λ ∼= N ⊕ Λ =⇒ M ∼= N.

Observation 1.3.12. If Λ has the cancellation property, then the class of a locally free module M is trivial in the classgroup if and only if M is free. Theorem 1.3.13. Let G be a nite group. The order Z[G] in the Q-algebra Q[G] has the cancellation property whenever one of the following holds:

• Gis abelian • Ghas order odd

Proof. See [25, Theorem 9.9].

In some cases the structure of Cl(Z[G]) is known; for example:

• Cl(Z[G]) = 0 if G is cyclic of order n, with 1 ≤ n ≤ 11 (see [6, p.253]). • Cl(Z[G]) has order 2 if G = H8(see [10, p. 48]).

Anyway, the structure of Cl(Z[G]) is unknown for almost every G.

Denition 1.3.14. A module M over a ring A is projective if for every surjec-tive module homomorphism f : N → M there exists a module homomorphism h : M → N such that f ◦ h = IdM.

Proposition 1.3.15. A module P is projective if and only if there exists a free module F and another module Q such

P ⊕ Q ∼= F.

In particular a nitely generated module is projective if and only if it is direct summands of a free nitely generated module.

Proof. See [15, Chapter 3,§4].

Proposition 1.3.16. A module over OK[G] is locally free if and only it if is

projective.

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Observation 1.3.17. Every element [M − N] in Cl(OK[G])is in the class of

a locally free module. In fact, for the previous propositions, we know that for every locally free module N there exists a locally free module N0 such that

N ⊕ N0 is free. Then,

[M − N ] = [M ⊕ N0] and M ⊕ N0 is a locally free module. Then,

Cl(OK[G]) ∼=

n

M locally free modules over OK[G]

o 

∼ with the equivalence relation

M ∼ N ⇔ ∃n, m ∈ N s.t. MMOK[G]m∼= N

M

OK[G]n.

Theorem 1.3.18. Let M be a OK[G] locally free module. If we see M as a

Z[G]-module, then it is locally free over Z[G]. Proof. There exists an OK[G]-module Q such that

Q ⊕ M ∼=OK[G]OK[G]

r,

because M is projective. Since OK is a free Z-module, then OK[G]is free over

Z[G] (we just need to take the same basis). If we see M and Q as Z[G]-modules, then

Q ⊕ M ∼=Z[G]OK[G]r,

that is free over Z[G]. Then, M is Z[G] projective and this concludes the proof.

1.4 Hom-description

In this section we want to give a dierent description of the classgroup, the so-called Hom-description (see Theorem 1.4.7). We will follow [10, I.§2]. We shall introduce the concept of determinant of a representation. Let Γ be a nite group and let T be a complex representation of the group Γ of degree n. Let K be a number eld and dene

ΩK :=Gal(Q/K).

Denition 1.4.1. We denote by RΓthe group of virtual characters of a group Γ,

namely the free group generated by the characters of irreducible representations of Γ.

Thanks to the fact that the sum of two characters of a representation is a character of a representation, it follow easily that every element of this group can be written as χ1− χ2 with χ1 and χ2 two characters.

Denition 1.4.2. Let χ be a character associated to an action ρ of Γ over a complex vector space V . Then the degree of χ is the dimension of V . Let χ = χ1− χ2 ∈ RΓ, where χ1 and χ2 are two characters. Then we dene the

degree of χ as

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Theorem 1.4.3. Let T be a complex representation of Γ of degree n. We can suppose that T takes values in Mn(Q).

Proof. This is an easy corollary of Brauer induction theorem, which can be nd in [23, Theorem 2.1.20].

If C is a commutative algebra over OK, then we can extend T to C[Γ]

T : C[Γ] 7→ C ⊗OKMn(Q) ∼= Mn(C ⊗OKQ) by setting X γ∈G cγγ 7→ X γ∈G cγ⊗ T (γ).

We can extend again the representation with the map

T : Mm(C[Γ]) 7→ Mm(Mn(C ⊗OKQ))∼= Mmn(C ⊗OKQ)

applying the previous map T to the coecients of the matrix. Since the map is an homomorphism, we have that T maps invertible elements in invertible elements and thus we have

T : GLm(C[Γ]) 7→ GLmn(C ⊗OKQ).

Composing with the determinant we obtain

detT : GLm(C[Γ]) 7→ (C ⊗OKQ)

.

Therefore, we can dene a map

detχ: GLm(C[Γ]) 7→ (C ⊗OKQ)

that sends

M → detT(M )

with detχ = detT for T a representation of the character χ. For the

representa-tion theory detχ is well-dened, namely the denition does not depends on the

choice of T .

Moreover, if χ is in RΓ, then we can write χ = χ1− χ2with χ1and χ2character

and so, for every χ ∈ RΓ, we dene

detχ=

detχ1

detχ2

. Denition 1.4.4. Let M ∈ GLm(C[Γ]). We dene

det(M ) : RΓ7→ (C ⊗OKQ)

by setting

det(M )(χ) = detχ(M ).

Let σ ∈ ΩQ. There is a natural action of σ on a character χ dened by

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Furthermore, σ can act on (C ⊗OKQ)

with

σ(a ⊗ b) = a ⊗ σ(b). Thus if M ∈ GLm(C[Γ])then

det(M )(χσ) = σ(det(M )(χ)).

From the denitions it is easy to check that detχ1+χ2(M ) = detχ1(M ) detχ2(M )

and then we conclude that

det(M ) ∈HomΩQ(RΓ, (C ⊗OKQ)

).

Since Γ is a nite group, we can choose a number eld E, depending on Γ, such that

det(M ) ∈HomΩQ(RΓ, (C ⊗OKE)

).

In fact we just need to take E such that the image of all the representations of Γhave the coecients in E and this is possible since Γ is nite and for Theorem 1.4.3. Mm(OK[Γ]) is an order of the K-algebra Mm(K[Γ]), then we know how

it is dened J(Mm(K[G]), Mm(OK[Γ])). Since it holds that:

(Mm(K[Γ]) ⊗KKP)∗∼= GLm(KP[Γ]) and (Mm(OK[Γ]) ⊗OK(OKP)) ∗= GL m(OKP[Γ]) we have J (Mm(K[G]), Mm(OK[Γ])) ⊆ Y P GLm(KP[Γ]) and α ∈ J(Mm(OK[Γ]))if αP ∈ GLm(OKP[Γ]) a.e.

If we take λ ∈ J(Mm(K[G]), Mm(OK[Γ])), then its P -th component λP is in

the group GLm(KP[Γ]). Taking the determinant of λP as we did before we have

a map

det(λP) : RΓ7→ (KP⊗OKE)

and then det(λP)has image in EP∗, where EP =QpEp with the product that

runs over all the prime in E that divides P (recall that KP ⊗OK E ∼= EP,

see [22, 6.1.1]). Over EP we can build an action of ΩK. Take γ ∈ ΩK and we

indicate with

iP : E 7→ EP

the canonical inclusion. We can consider

γ0 : iP(E) 7→ iγ(P )(E)

given by

γ0(x) = iγ(P )◦ γ ◦ i−1P (x).

The map γ0is obviously continuous and thus it admits a unique extension to the

whole eld EP. By the construction it is clear that the map is an isomorphism

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detm: J (Mm(OK[Γ])) 7→HomΩK(RΓ, J (E))

by setting

(λP)P 7→ (det(λP))P.

For more details about this construction see [10]. We want now to build an homomorphism

det :Cl(Λ) 7→ HomΩK(RΓ, J (E))

where Λ is an OK-order in K[Γ]. Let M be a locally free module. We know

that MK and MP are free of rank m. Let v1, . . . vmbe a basis of MK over K[G]

and w1,P, . . . wm,P basis of MP over ΛP, for every P . It is clear that

MK⊗OKKP ∼= KP[Γ]

m

and

MP⊗OKK ∼= KP[Γ]

m.

Then, vi⊗ 1and wi,P ⊗ 1 are both bases of KP[Γ]m over KP[Γ]. Then, there

exists a unique invertible matrix λP ∈ GLm(KP[Γ])moving the rst basis into

the second one. Set now (λ) = (λP)P. It is true that λ ∈ J(Mm(OK)[Γ])

(see [10, II.1]) and so we can dene

det(M ) = detm(λ).

Denition 1.4.5. Let us dene the group homomorphism det :Cl(Λ) 7→ HomΩK(RΓ, J (E))

such that, for every locally free module, M → det(M ).

Now we are ready to give the main results of this section. The proof of these theorems is far from being immediate and requires a great deal of eort. For the details see [10].

Theorem 1.4.6. The map det is well dened, i.e. it is independent form the choice of the bases.

Theorem 1.4.7. Let us denote with π the projection π :HomΩK(RΓ, J (E)) 7→ HomΩK(RΓ, J (E)) HomΩK(RΓ, E ∗)(det 1(U (Λ))) . Then, π ◦ det is an isomorphism and therefore

Cl(Λ) ∼= HomΩK(RΓ, J (E))

HomΩK(RΓ, E

) det

1(U (Λ))

. This isomorphism will be very useful in this work.

It is known that a locally free module M over OK[G] it is also a locally free

module over Z[G] and then we would like to know what is the element that represent M in the classgroup of Z[G], under the map of the Hom-description. We want to dene the norm of an homomorphism. In order to do that we need some preliminaries denitions. Let G be a group.

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Denition 1.4.8. Let H be a subgroup of G. A right coset of H is a non-empty subset S ⊆ G such that there exists a g ∈ G with S = gH.

Denition 1.4.9. A subset T of elements of G is called a right transversal of H if T contains exactly one element of each right coset of H.

Denition 1.4.10. If K is a number eld, then the norm map is NK/Q:HomΩK(RG, J (E)) →HomΩQ(RG, J (E))

dened by

NK/Q(f (χ)) =Yf (χσ−1)σ

where the product runs over a set of right transversal of ΩK in ΩQ. The

deni-tion is independent from the choice of set of right transversal (see [10, I.§2]). Theorem 1.4.11. The following diagram is commutative

Cl(OK[G]) −−−−→ HomΩK (RG,J (E)) HomΩK(RG,E∗)det1(U (OK[G]))   y   yNK/Q Cl(Z[G]) −−−−→ HomΩQ(RG,J (E)) HomΩQ(RG,E∗)det1(U (Z[G]))

where the rows are the isomorphisms of Theorem 1.4.7, the right hand column is induced by NK/Qand the left hand column is a map that see a OK[G]-module

as a Z[G]-module.

Proof. See [10, Theorem 2].

1.5 Tate's cohomology

In this section we shall introduce the Tate's cohomology, which will be very useful later. We will use some denitions and results of homological algebra, for more details see [29]. First of all, we will recall the classical denitions of cohomology and homology of a module over a group, then we will dene the Tate's cohomology and, in the end, we will prove some useful results. We will follow [20, Part 3].

Let G be a nite group and let A, B be G-modules. If f : A 7→ B is a G homomorphism, then f maps AG in BG. Consider an exact sequence of

G-modules

0 −→ A −→ B −→ C → 0, then the sequence

0 −→ AG−→ BG−→ CG

is also exact, so AG is a right derived functor. By denition, the right derived

functors of the functor AG are the cohomology groups of G with coecients

in A. We denote them with Hq(G, A). We can give an equivalent denition.

Consider Z as a G-module with the trivial action and consider a projective resolution of Z, i.e. an exact sequence

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where Pi are G-modules projective. Then, putting Ki=HomG(Pi, A)we have

a chain complex K and

Hq(G, A) = Hq(K).

This is the cohomology of the G-module A. We want to dene also the homology. Let DA be the subgroup of A generated by sa − a with s ∈ G and a ∈ A. We will denote the quotient A/DA with AG. It is easy to prove that functor AGis

right exact. Its left derived functors are, by denition, the homology group of Gwith coecients in A, denoted by Hq(G, A). Observe that H0(G, A) = AG

and H0(G, A) = AG. There is a natural map N from H0(G, A) to H0(G, A),

that sends the class of a to P ga.

Denition 1.5.1. The Tate's cohomology groupsHeq(G, A)are dened by: • eHn(G, A) = Hn(G, A)if n ≥ 1

• eH0(G, A) =Coker(N)

• eH−1(G, A) =Ker(N)

• eH−n(G, A) = Hn−1(G, A) if n ≥ 2.

Now, we shall study some properties of the Tate's cohomology.

Proposition 1.5.2. Let A be a G-module. If g is the order of G, then g eHq(G, A) = 0

for every q ∈ Z.

Proof. Let f : G0 7→ G be a homomorphism of groups. If A is a G-module,

then A has also a structure of G0-module, given by g0a = f (g0)a. We denote this

G0-module with f∗A. Since AG⊆ (f∗A)G0, we can take the inclusion map from

H0(G, A) into H0(G0, fA) and for the universal property of derived functors

we can extend this map to a map from Hq(G, A) into Hq(G0, fA). If H is a

subgroup of G and f is the inclusion from H to G, then the map dened below is called restriction. If H is a normal subgroup and we put

NG/H(a) =

X

g∈G/H

ga,

then we have a map from H0(H, A) to H0(G, A). We can extend it to a map

from Hq(H, A) to Hq(G, A)and this map is called corestriction.

Proposition 1.5.3. If G is a nite group and n =Card(G/H), then Cor ◦ Res = n,

where Cor is the corestriction and Res the restriction. Proof. See [20, p. 138].

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As for the cohomology we dene two maps, Cor and Res, also for the homology. Taking a map f : G0 → Gwe can dene a map from H

q(G0, f∗A) in Hq(G, A)

as we did for the cohomology, namely, by extending the inclusion of (f∗A) G0

in AG. If the map is the inclusion of the subgroup H in G, then we have the

corestriction map Hq(H, A) → Hq(G, A). If H is a normal subgroup of G we

dene NG/H0 : AG→ AH by NG/H0 (a) = X g∈G/H g−1a

and we can extend this map to a map, called Res, from Hq(G, A)to Hq(H, A).

As for the cohomology the composition of the restriction with the corestriction gives the multiplication by n (see [20, p. 128]). We can easily extend the denitions of Res and Cor to Tate's cohomology. So, we have two maps, Res and Cor, such that

Cor ◦ Res :Heq(G, A) → eHq(G, A) with

Cor ◦ Res = n Id

where n = [G : H]. Taking as H the subgroup of order 1 we have that n = g; moreover, it is easy to check that Heq(H, A) = 0 for every q. In this case the corestriction is the trivial map from a group of order 1 and then, for every q, we have that

g eHq(G, A) = 0.

Remark 1.5.4. In this thesis we will always work with Tate's cohomology and then we will write H instead ofHe, for the Tate's cohomology.

Denition 1.5.5. A G-module A is said cohomologically trivial if, for every subgroup H of G, Hq(H, A)is trivial.

Theorem 1.5.6. Let A be the ring of integers of a number eld, G a nite group and M an A[G]-module nitely generated. Then, M is G cohomologically trivial if and only if there exists a locally free resolution of A[G]-modules, i.e. an exact sequence

0 → P0→ P1→ M → 0

with P0and P1 locally free.

Proof. See [3, p. 457].

Corollary 1.5.7. Let A be the ring of integers of a number eld and M an A[G]-module nitely generated with G a nite group. If M is locally free over A[G], then it is G cohomologically trivial.

Proof. If M is locally free, then the exact sequence 0 → 0 → M → M → 0

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1.6 Square root of the inverse dierent

In the previous sections we dened the classgroup and the instruments involved in its study; now we are ready to dene the main subject of this thesis. We will dene the square root of the inverse dierent of a Galois extension and we will try to understand when it is locally free over the Z-order Z[G] in the algebra Q[G], with G the Galois group of the extension.

Let N/E be a Galois extension of number elds with Galois group G. Using the action of G on N we have that N is a Z[G]-module with the multiplication

(Xaγγ)x =

X aγγ(x).

Denition 1.6.1. The set

CN/E= {x ∈ N |TrN/E(xON) ⊂ OE}.

is called the inverse dierent of the extension.

Proposition 1.6.2. CN/E is a fractional ideal, i.e. a ON-module contained in

N with the property that there exists d ∈ ON such that dI ⊆ ON.

Proof. Let

d = det((TrN/E(wiwj))i,j),

where wi is a basis N over E, with wi∈ ON. We want to prove that

dCN/E⊆ ON.

This would conclude the proof since d ∈ ON and CN/E is a ON-module. Let

x ∈ CN/E, then

x =Xaiwi

with ai∈ E. We know that

TrN/E(xwj) ⊆ OE

and

TrN/E(xwj) =

X

aiTr(wiwj).

Using the Cramer's rule, we obtain ai=

det(Ai)

d

where (Ai)is a matrix with coecients in OE, since wi∈ ON and Tr(xwi) ∈ OE.

Then, dai∈ OE and so dx ∈ ON.

Proposition 1.6.3. Every fractional ideal I in ON can be written as

I =YPai

i

where the product runs over a set of nite primes Pi of ON and ai∈ Z.

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Thanks to the previous proposition we have that CN/E∼=

Y Pai

i

with ai∈ Z and Piprimes of ON. The inverse dierent CN/E is a square if and

only if ai are even for every i.

Denition 1.6.4. If CN/E is a square, then we can consider the square root of

the inverse dierent and we will denote it with A(N/E).

Observation 1.6.5. We just give the denition of the square root of the inverse dierent in the case of an extension of number elds. Actually, we can extend this denition to local number elds. Let E be a nite extension of Qp and N

be a nite extension of N. It appears clear that, if N/E is a Galois extension, then we can dene the square root of the inverse dierent also in this case, in the same way.

Denition 1.6.6. Let N/E be a Galois extension of number elds. Let Q ⊂ OE

be a prime ideal and it is well known that QON is an ideal of ON. Consider

the decomposition in prime of QON,

QON =

Y Pai

i .

We dene ai as the ramication index of Pi over Q and we will denote it with

e(Qi|P ).

Denition 1.6.7. Let Q ⊂ OE be a prime ideal and let

QON =

Y

Pe(Pi|Q)

i

be its factorization into primes. We say that a prime P is tame on Q if p := Q∩Z does not divide e(P |Q). If p divides e(P |Q), then the prime Q is said wild on P. The extension is said tame if every prime P is tame on Q := P ∩ OE.

Proposition 1.6.8. Let N/E be a Galois extension of number elds with Galois group G. Then, the following conditions are equivalent:

• N/E is tame. • TrN/E(ON) = OE.

• ON is OE[G]-projective.

Proof. See [10, Corollary 2,§ 3,I].

Denition 1.6.9. Let N/E be a Galois extension of number elds and P a prime of ON. The i-th ramication group for a prime P is the group

Gi(P, N/E) =

n

σ ∈Gal(N/E) | σ(α) − α ≡ 0 mod Pi+1∀α ∈ ON

o . Lemma 1.6.10. Let p = P ∩ Z. The order of G0(P, N/E)is equal to e(P |Q),

where Q = P ∩ OE. The rst ramication group G1(P, N/E) is a p-Sylow of

G0(P, N/E). Furthermore, for every i ≥ 1, the quotient

Gi(P, N/E)/Gi+1(P, N/E)

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Proof. See [20, p. 75].

Denition 1.6.11. The Galois extension N/E is weakly ramied if all the second ramication group attached to it are trivial.

Observation 1.6.12. If N/E is tame, then all the rst ramication groups are trivial. In particular the extension is weakly ramied.

Proposition 1.6.13 (Hilbert's formula). Let N/E be a Galois extension of number elds. Let G = Gal(N/E) and let C(N/E) be the inverse dierent. Let P be a prime of ON and let d = ordP(C(N/E)). Then

d = − ∞ X i=0 (ord(Gi(P, N/E)) − 1). Proof. See [20, p. 72].

The inverse dierent is a square if and only if ordP(C(N/E))is even for every

P. So, if the degree of the extension is odd, then ord(Gi(P, N/E))is odd for

every i and P . Using the Hilbert's formula we conclude, if the extension has odd degree, that ordP(C(N/E))is even for every P and then A(N/E) exists.

Theorem 1.6.14. Let N/E be an odd degree Galois extension of number elds. The following are equivalent:

1. A := A(N/E) is locally free over OE[G].

2. N/E is weakly ramied

Proof. Let P be a prime of N and V = G1(P, N/E). Let L be the subeld of

N xed by V , so Gal(N/L) = V and [N : L] = |G1| := pr. Furthermore, let

d = −ordP(C(N/L))and s = ordP(A).

1 =⇒ 2: If P is tame, then G1 is trivial and so is G2. So, we are interested

in the case when P is a wildly ramied prime. If A is locally free, then it is cohomologically trivial for 1.5.7 and H0(V, A) = 0. But

H0(V, A) = AV/TrN/L(A)

and this implies

TrN/L(A) = AV = A ∩ L.

To continue the proof we need two intermediate results.

Lemma 1.6.15. TrN/L(A) = A∩Lif and only if TrN/L(A(N/L)) = A(N/L)∩L

Proof. Since

A(N/E) = A(N/L)A(L/E), then

TrN/L(A) = A(L/E)TrN/L(A(N/L))

and

A ∩ L =A(N/L) ∩ LA(L/E). The proof follows easily.

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Lemma 1.6.16. Using the previous notation, TrN/L(A) = A ∩ Lif and only if

d = 2(pr− 1).

Proof. For [30, Proposition 4, Chapter 8], we know that hd + s

pr

i

=ordP(TrN/L(A)).

Then, if TrN/L(A) = A ∩ L, we have that

hd + s pr i =l s pr m = 1 +hs − 1 pr i ,

since N/L is totally ramied on P , where [m] represent the integer part of m and dme the smallest integer greater than or equal to m. We know that

|G1(P, N/L)| = pr, then d ≥ 2(pr− 1)and 1 +hs − 1 pr i =hd + s pr i ≥hs + 2p r− 2 pr i = 2 +hs − 2 pr i . Hence 1 ≤hs − 1 pr i −hs − 2 pr i

and writing s = t + kpr with 0 ≤ t ≤ pr− 1we conclude that t = 1 and then pr

divides s − 1. Then, hd + 1 pr i =hd + s + 1 − s pr i = 1 +hs − 1 pr i +h1 − s pr i = 1. But d ≥ 2(pr− 1)and then

d = 2(pr− 1), otherwise [d+1

pr ] would be greater than 1 . Vice versa, let us suppose that

d = 2(pr− 1). For every P in N, if we take P (L) = P ∩ L, we have

ordP (L)(TrN/L(A(N/L))) =

h d 2pr

i for [30, Proposition 4, Chapter 8] and

ordP (L)(A(N/L) ∩ L) =

l − d

2pr.

m We conclude easily, using the hypothesis d = 2(pr− 1), that

ordP (L)(A(N/L) ∩ L) =ordP (L)(TrN/L(A(N/L))).

Then it is easy to check that TrN/L(A(N/L)) = A(N/L)∩Land, for the previous

lemma,

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Now we have the instruments to conclude the proof of the rst implication of the theorem. We know that

TrN/L(A) = AV = A ∩ L

and clearly

|G0(N/L)| = |G1(N/L)| = pr.

So, thanks Lemma 1.6.16 and the Hilbert's formula, we have that G2(N/L)is

trivial. But

G2(N/E) = G2(N/E) ∩ G1(N/E) = G2(N/E) ∩Gal(N/L) = G2(N/L)

and then N/E is weakly ramied.

2 =⇒ 1If N/E is weakly ramied, then d = 2(pr− 1), so AV =Tr

N/L(A)and

hence

H0(V, A) = 0. Let us start proving that this implies that H0(G

p, A)for all Sylow group Gpof

G. Let Np be the eld xed by Gp. If all P in N above p are tamely ramied

in N/E, then N/Np is tamely ramied and so by [28, Theorem 1] we know

that H0(G

p, A) = 0. If P is not tamely ramied, let V = G1(P, N/E) and

L =Fix(N, V ). There exists a p-Sylow G0psuch that Fix(N, G0p) := Np0 ⊆ Land since L is normal and all the p-Sylow are conjugate we obtain that Np⊆ L. The

extension L/Npis tamely ramied and so it is easy to check, for every fractional

ideal I, that

TrL/Np(I) = I ∩ Np.

Then,

TrN/Np(A) =TrL/Np(TrN/L(A)) =TrN/L(A) ∩ Np.

Since we know that TrN/L(A) = A ∩ L, then TrN/Np(A) = A ∩ Np and so

H0(Gp, A) = 0.

Lemma 1.6.17. Any ambiguous ideal B (i.e. a fractional ideal with a structure of G-module) in N/E has the cohomology of a nite module.

Proof. By the normal basis theorem there exists c in ON such that N = E[G]c.

If we take b ∈ B ∩ E, then N = E[G]cb. Let us dene M = OE[G]cb, then M is

contained in B, since B is an ambiguous ideal and B/M is nite. We have an exact sequence

0 → M → B → B/M → 0.

Studying the long cohomology sequence and using that M is cohomologically trivial for 1.5.7 we deduce that the cohomology of B is the same of B/M. Lemma 1.6.18. Let G be a p-group. For a nite G-module M the condition H0(G, M ) = 0implies that M is cohomologically trivial.

Proof. See [12].

Lemma 1.6.19. A module M is G cohomologically trivial if and only if it is Gpcohomologically trivial for all Sylow subgroups Gpof G.

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Proof. See [20, IX,5].

Using the previous work we know that H0(G

p, A) = 0 for every Sylow. For

Lemma 1.6.17 we know that A has the same cohomology of M0, a nite module.

Then, M0 is G

p cohomologically trivial for every Sylow and, for Lemma 1.6.19,

we deduce that M0is cohomologically trivial. Since M0and A have the same

co-homology, we conclude that A is cohomologically trivial. If A is cohomologically trivial over G, then it is projective over OE[G] (because it is without torsion,

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Chapter 2

The odd degree case

Let N/E be a weakly ramied extension of number elds of odd degree. Denote by G its Galois group. We will show that A(N/E) is free over every maximal order in Q[G] which contains Z[G] and we will use this result to prove that A(N/E) is free over Z[G], when the extension is tamely ramied. We will follow [7] and we will use many results of [10]. In our hypothesis the extension has degree odd and it is weakly ramied, hence A(N/E) always exists and it is locally free over OE[G].

2.1 Localization to a maximal order

Proposition 2.1.1. The rank of A(N/E) over OE[G]is 1.

Proof. It is enough to prove that

A(N/E) ⊗OEE ∼= N,

since N ∼= E[G](see Proposition 1.3.2). It is easy to show that every element in A(N/E) ⊗OEE can be written in the form x ⊗ (1/d) with x ∈ A(N/E) and

d ∈ Z. Consider the map

A(N/E) ⊗OEE → N

dened by

(x ⊗ y) → xy;

we want to prove that this is an isomorphism. Since every element z in N can be written as z = z0/mwith z0∈ O

N ⊆ A(N/E) and m ∈ Z, then the image of

(z0⊗ 1/m) is z and then the map is surjective. For the injectivity suppose that there exist x, x0∈ A(N/E)and m, m0 ∈ Z such that the image of (x ⊗ 1/m) is

equal to the image of (x0⊗ 1/m0), i.e. x/m = x0/m0. Then,

(x ⊗ 1 m) = (xm 0 1 mm0) = (x 0m ⊗ 1 m0m) = (x 0 1 m0).

Hence, the map is injective.

Let χ be a character of G and c in N. 21

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Denition 2.1.2. The resolvent is dened by (c|χ) = det X

g∈G

g(c)T (g)−1, where T is a representation with character χ.

In Section 1.4 we dened the determinant of a representation. Using that de-nition we have

detχ(g) = det(T (g)),

where T is a representation of G with character χ.

Let c ∈ N be such that N = E[G]c and for every prime P in E let bP ∈ A(N/E) ⊗OEOEP = (A(N/E))P

such that

(A(N/E))P = (OEP[G])bP.

Moreover, let L be a number eld "big enough", namely an extension of N which also contains the image of all characters of G, the values of arithmetical functions that will occur, e.g. resolvent and Gauss sums, and such that the algebra Q[G] splits. For every χ is well dened (b|χ) in J(L) as (bP|χ)P (see [10, 1.4]). Let

f ∈HomΩQ(RG, J (L))such that

f (χ) = NE/Q((b|χ)(c|χ)−1).

Theorem 2.1.3. The class of A(N/E) in Cl(Z[G]) corresponds to f in the Hom-description.

Proof. We know that the class of A(N/E) in Cl(OE[G]) corresponds, in the

Hom-description, to a map v such that

v(χ)P = detχ(λP)

with λP ∈ EP[G]and λPc = bP. Furthermore, if

λP =

X

g∈G

σgg,

and T is a representation with character χ, then X g∈G g(λPc)T (g)−1= X g,γ∈G g(γ(c))σγT (gγ)−1T (γ) = X g∈G g(c)T (g)−1 X γ σγT (γ)  and applying the determinant we obtain

(bP|χ) = (λPc|χ) = (c|χ) detχ(λP).

Then, the class of A(N/E) in Cl(OE[G]) corresponds, in the Hom-description,

to the map

v(χ)P = (bP|χ)(c|χ)−1

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Now we dene the Gauss-Galois sum. Let F be a nite Galois extension of Qp

and θ be a multiplicative homomorphism θ : F∗→ C∗

of nite order. Let Um(F ) = 1+PFmwhere PFis the maximal prime of F and let

U0(F ) = OF. Then, there exists a minimal integer m such that Um(F ) ⊆ ker(θ)

and let us dene the conductor as f(θ) = Pm

F . Let ψpbe the additive character

ψp : Qp → C∗ such that ψp(Zp) = 1and ψp(1/pr) = e2πi/p

r

. Composing this map with TrF /Qp we obtain

ψF := ψp◦TrF /Qp.

Let D(F ) := C(F/Qp)−1 be the dierent of F/Qpand choose c ∈ F∗ such that

(c) = f (θ)D(F ). Then, the Gauss sum is dened as τ (θ) =X

u

θ(uc−1)ψF(uc−1),

with u running through a complete set of representatives of U(F ) modulo 1 + f (θ). It is easy to check that τ is well-dened, i.e. does not depend from the choice of u or c.

Theorem 2.1.4. Let F be a nite Galois extension of Qp.Then, for all nite

extension E/F with Galois group G, there exists an homomorphism χ → τ (χ, E/F )

of RG into Q ∗

with the following properties.

• Let χ ∈ RG abelian (i.e., a character of degree 1). The composition with

the Artin map (for the denition of this map see [17, Theorem 3.4]) gives us a multiplicative character θ of E. Then τ(E/F, χ) = τ(θ), where τ(θ) is the Gauss sum dened above.

• If F ⊆ E ⊆ E0, χ ∈ R

G and deg(χ) = 0, then

τ (E, χ) = τ (E0,IndGG0χ),

where G = Gal(E/F ), G0=Gal(E0/F )and the degree of χ is dened as

in 1.4.2.

Proof. See [10, Theorem 18].

Denition 2.1.5. Let N/E be an extension of number elds with Galois group G, χ ∈ RG and let us dene the Galois Gauss sum as

τ (N/E, χ) =Y

P

τ (NP/EP ∩E, χDP)

where the product runs over all primes of N and χDP is the restriction of χ to

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Lemma 2.1.6. Let us dene the Adams operation over RG as the map

ψ(χ)(g) = χ(g2). The map ψ:

• has image in RG;

• commutes with induction, restriction, ination and with the action of Ω(Q) on RG;

• is such that ψ(detχ) = detψ(χ).

Proof. The rst sentence holds because irreducible characters are a free basis of the group of maps that are constant on each conjugacy classes and

ψ(χ)(hgh−1) = χ(hg2h−1) = χ(g2) = ψ(χ)(g).

For the second one see [14]. For the last one, let us take F , the family of all cyclic subgroups of G. The restriction gives us an injective homomorphism

RG ,→

Y

C∈F

RC.

Then, we just need to prove the sentence in the cyclic case and this is easy since the irreducible cyclic characters are of degree 1 and then ψ(χ) = χ2.

Theorem 2.1.7. The class of A(N/E) ∈ Cl(Z[G]) is represented, in the Hom-description, by the homomorphism

vN/E(χ) = NE/Q(b|χ)τ (N/E, ψ(χ) − χ)−1.

Proof. If we prove that

NE/Q(c|χ)τ (N/E, ψ(χ) − χ)−1 is ΩQ invariant, then it is in HomΩQ(RG, E

) and then v and f represent the

same class. Thanks to [10, Theorem 20A-B] we can conclude. In the next pages we will prove

Theorem 2.1.8. If N/E is an odd degree extension weakly ramied, then v ∈HomΩQ(RG, U (L)).

Theorem 2.1.9. If the extension has odd degree, is weakly ramied and M is a maximal order that contains Z[G], then

M⊗Z[G]A(N/E) is free over M.

Proof. If M is a maximal order, then

HomΩQ(RG, U (L)) = det(U (M))

for [10, I.2.19]. Hence, we conclude, using the Hom-description and Theorem 2.1.8, that the class of M ⊗Z[G]A(N/E) is trivial in Cl(M). But M has the

cancellation property (see [25, Theorem 9.9.]) and then for Observation 1.3.12 we conclude that A(N/E) is free.

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Now, we want to generalize the denition of the norm given in Chapter 1. Let M be a nite Galois extension of Qp and let M0 be a nite Galois extension of M.

Let ΩM =Gal(Qp/M )and ΩM0 =Gal(Qp/M0). Let K and H be ΩM-modules.

We shall consider the norm map NM0/M :Hom M 0(K, H) →HomΩM(K, H) dened by NM0/Mf (k) = Y f (kσ−1)σ

for f ∈ HomΩM 0(K, H)and k ∈ K. Here σ runs over a right transversal of Ω0M

in ΩM.

The proof of Theorem 2.1.8 is very technical, so we will not give every details. The most important step in the proof is to show the following:

Theorem 2.1.10. Let N/E be a totally ramied extension of odd degree e and containing Qp. Let us suppose that A(N/E) is free over OE[G] and let a be

a free generator of A(N/E) over OE[G]. Then, using the notation introduced

above, it holds the following equation of ideals

(NE/Qp(a|χ)) = (τ (ψ(χ) − χ))

for every local character χ of G, i.e. a character of G composed with a xed embedding j : Q → Qp.

In the next two sections we will prove this theorem. Let us start introducing some notation.

Denition 2.1.11. Let N/E be a totally ramied extension of odd degree e and containing Qp. Let G = Gal(N/E) and let us suppose that A(N/E) is free

over OE[G]. Let a be a free generator of A(N/E) over OE[G]. Then, we dene

R(χ) := (a|χ) for every local character of G.

Since a totally ramied extension of local elds can be either tame or wild, we will prove the theorem in this two cases.

2.2 Proof of Theorem 2.1.10 in tame case

In this section we will prove

Theorem 2.2.1. Let N/E be a totally and tamely ramied extension of odd degree e and containing Qp. Let us suppose that A(N/E) is free over OE[G].

Then, using the notation introduced above,

(NE/QpR(χ)) = (τ (ψ(χ) − χ)) for every local character χ of G.

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It is well know that the extension is cyclic. Let us take an uniformizer πN such

that πE = πNe is an uniformizer for E and N = E(πN). Recall that, since the

extension is totally ramied, then the e-th roots of unit are in E. Let G be the Galois group of the extension. If we denote

θ(g) = g(πN)/πN,

then every local character can be written as χ = θν(χ) with 0 ≤ ν(χ) < e.

Lemma 2.2.2. Using that ON = OE[πN], we deduce that

b = 1 + πN + . . . πNe−1

is a free generator of ON over OE[G].

Proof. We want to prove that ai= {gi(b)}0≤i≤e−1 is a OE-basis of ON, with

ga generator of G. We know that g acts on πN as the product of ω, a primitive

root of unit, and then, if we call Π the vector Π = (πi

N)i and a the vector

a = (ai)i, we have that

a = AΠ

with A a matrix with Ai,j = ωij. This is a Vandermonde's matrix and then it

is well known that it determinant is Y

i≤j

(ωi− ωj).

If we prove that every element of the product is a unit, then using the Cramer's rule we have

πi= det(Ai) det(A)

with det(Ai) =P zJaJ with zJ units of OE and then we conclude the proof.

Since ωi∈ P/

E, the maximal prime of OE, then we just need to check that

ωi− 1 /∈ PE

for 1 ≤ i ≤ e − 1. Let us suppose that ω − 1 ∈ PE (if it is ωi− 1 ∈ PE with

i 6= 1the proof is very similar). Then, multiplying for ωe−1 we obtain that

ωe−1− 1 ∈ PE.

Then, ωe−1− ω ∈ P

E and thus ωe−2− 1 ∈ PE. Repeating this method we can

prove that

ωi− 1 ∈ PE

for every i and then

e = e − e−1 X i=0 ωi= e−1 X i=0 1 − ωi∈ PE.

This is absurd because the extension is tame, so ωi− 1 /∈ P

E and (ai) is a

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If I = (πN)ν(I) is any fractional ideal of N, then we can take a free generator

of I over OE[G]taking

a = bπNν(I).

Lemma 2.2.3. Let s(χ) = ν(χ)/e and s(I) = ν(I)/e. Then, (a|χ)OE= (πE)s(χ)−[s(χ)−s(I)] and (b|χ)OE= (πE)s(χ). Proof. We have (a|χ) =X g∈G e−1 X i=0 g(πN)ν(I)+iχ(g−1) =X i πNν(I)+iX g θ(g)ν(I)+i−ν(χ) = πNν(χ)X i πNν(I)+i−ν(χ)X g θ(g)ν(I)+i−ν(χ) = eπNν(χ)πnN where n = ν(I) − ν(χ) + (ν(χ) − ν(I))0

with m0 is the remainder of the division of the integer m by e. Using that

πe

N = πE we conclude. The second equation follows from

(b|χ) =X i,g g(πN)iχ(g)−1 = X i πNi X g θ(g)i−v(χ)= πv(χ)N e.

Let f be the integer such that NE/Qp(πE) = (p)

f.

Lemma 2.2.4. Using the previous notation we have that (NE/Qp((a|χ)(b|χ)−1)) = (p)s with s = f −1 X j=0 ([pjs(χ)] − [pjs(χ) − s(I)]) and (τ (χ2− 2χ)) = (p)q with q = f −1 X j=0 (2[pjs(χ)] − [2pjs(χ)]). Furthermore, if we put I = A(N/E), then s = q and

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Proof. For the rst sentence let us dene Q(χ) the ideal ((a|χ)(b|χ)−1). In

order to calculate the Norm of Q(χ) we have to take a set of transversal of ΩE over ΩQp. Let us take M the maximal unramied subextension, then f =

[M : Qp]. If we take ω the Frobenius of M/Qp, then the transversals are in the

form σωj with j less than f and σ runs over the transversals of Ω

F over ΩM. For [10, Theorem 25], Q(χωjσ−1) = Q(χpj). Observe that ν(χk) = kν(χ) − e[ν(χ)k e ] and then s(χk) = k(s(χ)) − [s(χ)k]. Hence, for the previous lemma,

Q(χωjσ−1) = Q(χpj) = (πE)b(j) with b(j) = −[pjs(χ)] − [pjs(χ) − s(I)]. Then, (NE/QpQ(χ)) =Y j,σ Q(χωjσ−1)σω−j =Y j,σ Q(χpj)σω−j =Y j (πE)eb(j)= (p)s.

For the second sentence see [10, Theorem 27]. For the last one let I = A(N/E) and so

s(I) = −(e − 1)/2e.

Fix j < e and put c = c(j) = pjs(χ) and d = −s(I). Then, we want to prove

that 2[c] − [2c] = [c] − [c + d]. With some easy calculation we have that 2[c] − [2c] = ( 0if c < [c] + 1/2 −1if [c] + 1/2 ≤ c and [c] − [c + d] = ( 0if c + d < [c] + 1 −1if [c] + 1 ≤ c + d. Before to conclude we need an auxiliary lemma:

Lemma 2.2.5. Let c = pjν(χ)/eand d = (e − 1)/2e. Then, [c] + 1/2 ≤ c if and

only if [c] + 1 ≤ c + d.

Proof. With some easy calculation

[c] + 1/2 ≤ c ⇔ [c] + 1 ≤ c + d + 1/2 − d ⇔ [c] + 1 ≤ c + d + 1/2e. Then, we can conclude easily using the fact that e is odd.

Using the previous lemma we have s = q and then

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In conclusion, using that τ(χ) and NE/Qp(b|χ) only diers by unit (see [10, Theorem 25]), then (τ (ψ(χ) − χ) = (τ (χ))(τ (χ2− 2χ)) = (NE/Qp((a|χ)(b|χ)−1))(τ (χ)) = (NE/Qp(R(χ)))(NE/Qp((b|χ)−1))(τ (χ)) = (NE/Qp(R(χ)))

and then we have proved the Theorem 2.2.1.

2.3 Proof of Theorem 2.1.10 in wild case

In this section we will prove

Theorem 2.3.1. Let N/E be an extension of local elds containing Qp. Let

us suppose that N/E is wildly ramied and let G be the Galois group of the extension. Let us suppose also that A(N/E) is free over OE[G]. We want to

prove that

(NE/QpR(χj)) = (τ (ψ(χj) − χj)). for every character χ of G and every embedding j : Q → Qp.

Let us recall some useful facts about the characters, for more details see [21]. First of all, if G is a group and if we denote with nithe degree of the irreducible

representations of G, then P n2

i = g, where the sum runs over all the irreducible

representations and g is the cardinality of G (see [21, p.31]). Furthermore, let H be a subgroup of a group G and χ a character of H, then the character of the induced representation is

(IndGHχ)(s) = 1 h X g∈G gsg−1 ∈H χ(gsg−1)

with h the cardinality of H. Let s ∈ G and dene Hs= H ∩ sHs−1. For every

x ∈ Hslet us dene χs(x) = χ(sxs−1)and Ressχ, as the restriction of χ on Hs.

Proposition 2.3.2 (Mackey's criterion). The induced character (IndG Hχ) is

irreducible if and only if χ is irreducible and, for every s ∈ G \ H, the represen-tations χsand Res

s(χ)are disjoint, i.e. < χs,Ress(χ) >Hs= 0.

Proof. See [21, 7.4].

Since N/E is totally wildly ramied, then we know that G = V o C, where C is cyclic of order that is not divided by p and V is an elementary abelian p−group (i.e V = (Z/pZ)k). For a proof of this see [20, Chapter 4, Corollary 4

of Proposition 7].

Lemma 2.3.3. If we take θ an irreducible character of V , then there is an action of C given by cθ(v) = θ(cvc−1). If θ is not trivial, then the orbit of the

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Proof. Let us take c ∈ C dierent from the identity and θ an irreducible character of V dierent from the trivial character. If cθ = θ, then for every v ∈ V we have that cvc−1v−1 belongs to ker(θ). But

ord(ker(θ)) = pr−1

and then there exists v0 6= e such that cv0c−1 = v0. The action of C over V

with the conjugation acts like the scalar multiplication on the Fp vector space

V (see [10, IV.2, Proposition 9]) and then it is absurd.

Lemma 2.3.4. Using the previous lemma we conclude that there are (pr− 1)/c

irreducible characters on G induced by the irreducible characters of V , namely θ0, . . . θ(pr−1)/c, one for every element in the orbit of the action of the previous

lemma.

Proof. Let θ be an irreducible non trivial character of V . We will use Mackey's criterion. Since V is normal, we just need to check that for every s ∈ G \ V we have

hθ, θsi V = 0.

If s = cw with w ∈ V and c ∈ C, then, using that V is abelian, we obtain X

v∈V

θ(v)θ(cwvw−1c−1) =X

v∈V

θ(v)(cθ)(v) = hθ, (cθ)iV = 0,

where the last equality follows from the fact that θ and cθ are two dierent irreducible characters of V . Now we want to check that IndG

Vθ =Ind G

Vθ0 if and

only if θ0 = cθ for some c ∈ C. Using the formula for induced characters we

have that IndG

Vθ(g) = 0if g /∈ V and IndG Vθ(v) = X c∈C cθ(v)

if v ∈ V . Then, using the fact that every character can be written uniquely as a sum of irreducible characters we obtain the proof.

Inating the characters of C we obtain ord(C) irreducible characters. Then, we have all the characters because

prc = c + c2(pr− 1)/c. If χ is inated from C, say χ = InfG

Cφ, then for every embedding j of Q into Qp

we have, for [10, Lemma III.1.5.],

NE/QpR(χj) = NE/QpR(φj) and

τ (χ) = τ (φ).

Moreover, Fix(N, V )/E is tamely ramied, hence for such a character χ we are done for the work in the previous section. Then, we have to consider the case of χ induced by a character θ of V .

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Lemma 2.3.5. Put H = ker(θ) and H00= \

c∈C

cHc−1.

Let G0 = G/H00, H0= H/H00and V0 = V /H00. We have

X

ω∈Gal(Q(ζp)/Q)

χω=IndGH00(1H0) − rC

where rC is the regular character of C lifted to G and IndG

0

H0(1H0)is a character

of G0 lifted to G.

Proof. Let k(g) be the cardinality of the set {c ∈ C|cgc−1 ∈ H}. We evaluate

both sides in g ∈ G:

rC(g) =

(

c if g ∈ V 0 otherwise and if we put g0 the class of g in G0, then we have

IndG0 H0(1H0)(g0) = X r∈G0 rg0 r−1 ∈H0 (1H0(rg0r−1)) = ( k(g)pif g0∈ V0 0otherwise. Furthermore X ω χω(g) = X {c:cgc−1∈V } X ω θω(cgc−1) = ( k(g)(p − 1) − (c − k(g))if g ∈ V0 0 otherwise

and this conclude the proof.

We have, for all ω ∈ ΩQ, the equalities of ideals

(NE/QpR(χωj)) = (NE/QpR(χj)) and

(τ (χω)j) = (τ (χ)j)

for [10, Theorem 20A-20B]. Let again a be a free generator of A(N/E) over OE[G]. Since the norm resolvent and the Gauss sum we are considering lie in a

same local eld, to prove Theorem 2.3.1 it is sucient to show the proposition. Proposition 2.3.6. The product over ω ∈ Gal(Q(ζp)/Q)

Y

ω

[NE/Qp(a|χωj)(τ (ψ(χω) − χω)−1)j] is a unit.

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Proof. It is easy to check that Y ω [NE/Qp(a|χωj)(τ (ψ(χω) − χω)−1)j] is equal to [NE/Qp(a X ω χωj)(τ (ψ(X ω χω) −X χ χω)−1)j]. Then, we just need to prove that

NE/Qpa Ind G0 H0(1H0) − rC  τ (ψ(IndGH00(1H0) − rC) −IndG 0 H0(1H0) − rC)−1 j is a unit. Since the order of G is odd, we obtain

τ (rC) = τ (ψ(rC)) and τ (ψ(IndGH001H0)) = τ (IndG 0 H0ψ(1H0)) = τ (IndG 0 H01H0). Then,  τ (ψ(X ω χω) −X χ χω)−1 j =τ (ψ(IndGH00(1H0) − rC) −IndG 0 H0(1H0) − rC)−1 j

is a unit. If we check that NE/Qp(a|rC)and NE/Qp(a|Ind

G0

H01H0)are units, then

we conclude. To continue the proof we need an intermediate result.

Lemma 2.3.7. Let L/E be a nite Galois extension with H = Gal(L/E). Then, for every b ∈ L, we have

(b|2rH) = det(Tr(γ(b)γ0(b))γ,γ0∈H).

Proof. We know that T , the regular representation of H, acts on the vector space generated by {eh}h∈H with T (g)eh= egh. Then, the matrix

A =X h h(b)T (h)−1 is such that Aeγ = X h h(b)eh−1γ = X h γ(h−1(b))eh.

Then, (A)h,γ= γ(h−1(b)). The coecient in position (γ, γ0)of ATAis

X

h

h−1(γ(b))h−1(γ0(b)) =Tr(γ(b)γ0(b)). Since

(b|2rH) = det(A)2= det(ATA),

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Let us dene L = Fix(N, V ). Since b := TrN/L(a)is a free generator of A(L/E)

we have that, with some easy calculation,

(a|2rC) = (b|2rG/V) = det(TrN/E(g(b)h(b)))g,h=discTr(A(L/E))

and that is a unit because A(L/E) is self-dual with respect to the trace form. So, to end the proof it is sucient to show that NE/Qp(a|Ind

G0

H01H0). Let us change

the notation. Let L = Fix(N, H00)containing M = Fix(N, H). As before we

can restrict our attention to the extension L/E, then we replace G0, H0 and

TrN/L(a) with G, H and a. Let also b be a free generator of A(L/E) over

OM[H]and let d(M/E) denote the discriminant of M/E. Using [10, Note 4 to

ChapterIII], we have the identity of ideals

((a|IndGH001)) = NM/E((b|1))d(M/E)1/2.

Now, (b|1) = TrL/M(b)and then

NM/E((b|1)) = NM/E(TrL/M(b))

= NM/E(A(L/E) ∩ M )

= NM/E(A(M/E))

= d(M/E)−1/2.

2.4 Conclusion of odd degree case

Now we are ready to conclude the proof of Theorem 2.1.10. Since a totally ramied extension of local elds can be either tame or wild, we conclude using Theorem 2.2.1 and 2.3.1.

Now that we have proved Theorem 2.1.10, we want to prove Theorem 2.1.8. Sketch of the proof of Theorem 2.1.8. As I said before, we will not give all details. We want to prove that

vN/E(χ) = NE/Q(b|χ)τ (N/E, ψ(χ) − χ)−1∈HomΩQ(RG, U (L)).

The rst step of the proof is to reduce the problem to the local case. Let P be a prime of E, let Q be a prime of N above P and p := P ∩ Z. Let us observe that A(NQ/EP) is free over OEP[Gal(NQ/EP)]. This fact can be proved repeating

the proof of Theorem 1.6.14. Let aQ be a free generator of A(NQ/EP) over

OEP[Gal(NQ/EP)]. If we prove that

(NEP/Qp(aQ|χ)) = (τ (NQ/EP, ψ(χ) − χ)

−1),

for every P and Q, then it follows the proof of Theorem 2.1.8. For the proof of this fact, see [7, Section 5]. Then we are reduce to prove the following

Theorem 2.4.1. Let N/E be an odd degree eld extension containing Qp.

Let us suppose that A(N/E) is free over OE[G]. Then, using the notation

introduced above,

(NE/QpR(χ)) = (τ (ψ(χ) − χ)) for every local character χ of G.

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To conclude the proof we need an intermediate result.

Theorem 2.4.2. Let N/E be an odd degree eld extension containing Qp. Let

us suppose that A(N/E) is free over OE[G]. Let B be an intermediate eld

which is non-ramied over E and we dene H = Gal(N/B). Then (τ (ResGH(χ))) = (τ (χ))

[B:E]

. and

NB/QpR(ResGHχ) = NE/QpR(χ)[B:E]. for any character of G.

Proof. For a proof of this fact see [7, Section 6].

It is clear that, if the last theorem holds, then the last result combined with the proof of Theorem 2.1.10 give us the proof of Theorem 2.1.8.

We conclude the chapter with a very important result, due to Erez.

Theorem 2.4.3. If the extension N/E is tame and has odd degree, then A(N/E)is free over Z[G].

Sketch of the proof. We will not prove this theorem, but we will give the main idea. The complete proof can be found in [7, Section 8]. We know by Theorem 2.1.8 that

v(χ) = NE/Q(a|χ)τ (N/E, ψ(χ) − χ)−1 lies in HomΩQ(RG, U (E)). If we prove that

v(χ) ∈HomΩQ(RG, (OE)

) det(U (Z[G]))

then we can conclude, using the Hom-Description, that the class of A(N/E) is trivial in Cl(Z[G]). Using 1.3.12 we conclude that A(N/E) is free.

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Chapter 3

The even degree case

Let N/E be a Galois extension of number elds with Galois group G. Let A(N/E)be the square root of the inverse dierent and suppose that it exists. The main goal of this chapter is to prove Theorem 3.1.4, which tell us that the class of A(N/E) is equal to the class of ON in Cl(Z[G]), if the extension

is locally abelian. Then we will use this result to try to understand when the class of A(N/E) is trivial. In order to do that we use a theorem of Taylor which give us an instrument to calculate the class of ON in ClZ[G]. In particular, we

will prove a theorem that shows, under some hypothesis, that A(N/E) can be trivial in the classgroup also in the even degree extensions. In the end we will nd an example of an extension with A(N/E) not trivial in the classgroup. We will follow [1].

3.1 Reduction to local case

In this chapter we will always assume that A(N/E) exists.

Denition 3.1.1. Let Q be a prime in ON. Let us dene the decomposition

group of Q as

DQ:= {σ ∈ G|σ(Q) = Q} < G.

Denition 3.1.2. Let P be a prime of OE and Q a prime over P in ON. The

extension NQ/EP has Galois group DQ. We say that the extension N/E is

locally abelian if DQ is abelian for every prime ideal Q.

Proposition 3.1.3. Let N/E be a tame G-Galois extension of number elds. Assume further that A(N/E) exists. Then, A(N/E) is locally free.

Proof. The proof in the odd case works also for the even case (see the proof of Theorem 1.6.14). The only dierence is that at some point in the proof one needs to observe that, if N/E is a weakly ramied Galois extension whose inverse dierent is a square, then the inverse dierent of N/L is also a square, where L = NV and V is a subgroup of Gal(N/E). This follows easily from

Hilbert's formula ( [20, IV, Proposition 4]). The rest of the proof goes through unchanged.

In the next pages we will prove the following: 35

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Theorem 3.1.4. Let N/E be a tame G-Galois extension of number elds. Assume further that N/E is locally abelian. Then,

(ON) = (A(N/E)).

Lemma 3.1.5. If

0 −−−−→ K −−−−→ P −−−−→ M −−−−→ 0φ and

0 −−−−→ K0 −−−−→ P0 −−−−→ M −−−−→ 0φ0

are short exact sequences of A-modules and P and P0 are projective, then

K ⊕ P0 ∼= K0⊕ P. Proof. Let X be the submodule of P ⊕ P0 dened by

X =n(p, p0)|φ(p) = φ0(p0)o.

The projection π from X to P is surjective since φ0 is surjective and then for

every p ∈ P there exists p0∈ P0 such that φ(p) = φ0(p0).The kernel of π is

ker(π) =n(0, p0) ∈ Xo ∼= ker(φ0) ∼= K0 and then the sequence

0 → K0→ X → P → 0 is exact. Since P is projective, we conclude that

X ∼= P ⊕ K0.

In the same way, using the projection π0 from X to P0, we can prove that

X ∼= P0⊕ K.

In the following pages we will study the module SN/E:= A(N/E)/ON.

In fact, thanks to the previous lemma, if we nd two locally free modules such that

0 → M → M0 → SN/E→ 0

is exact then, we obtain

M ⊕ A(N/E) ∼= M0⊕ ON.

So, if we show that (M) = (M0)in ClZ[G] then we can prove that

(A(N/E)) = (ON).

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